4(x-15)=0 mn giup em nha
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\(\left(x-8\right)\left(3x-15\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-8=0\\3x-15=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=5\end{cases}}\)
Vậy x = 5 hoặc x = 8
\(\left(x-8\right)\left(3x-15\right)=0\)
Hoặc \(x-8=0\Leftrightarrow x=8\)
Hoặc\(3x-15=0\Leftrightarrow x=5\)
Vậy \(x=8\)hoặc \(x=5\)
\(\left|x.\frac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)
\(\Rightarrow\left|x.\frac{4}{15}\right|-3,75=-2,15\)
\(\Rightarrow\left|x.\frac{4}{15}\right|=-2,15+3,75\)
\(\Rightarrow\left|x.\frac{4}{15}\right|=1,5\)
\(\Rightarrow\orbr{\begin{cases}x.\frac{4}{15}=1,5\\x.\frac{4}{15}=-1,5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{45}{8}\\x=-\frac{45}{8}\end{cases}}\)
Có phải theo cách của em là làm như này đúng ko ah Bảo Bình
| X+4/15| - |3,75| = -| -2,25|
=> |x+4/15| - 3,75 = -2.25
+ x+4/15> hoặc =0 =>|x+4/15|=x+4/15
=>x+4/15 - 3,75 = -2.25
x+4/15=1.5
x=45/8
+ x+4/15<0 => |x+4/15|=-(x+4/15)
=> -(x+4/15) - 3.75 = -2,25
-x+4/15=1.5
-x=48/5
=> x=-45/8
1) 5x = 120 : 2
5x = 60
x = 60 : 5
x = 12.
2) 3x + 27 = 45
3x = 45 - 27
3x = 18
x = 18 : 3
x = 6.
3) 73 + x - 122 = 234 : 2
73 + x - 122 = 117
73 + x = 117 + 122
73 + x = 239
x = 239 - 73
x = 166.
4) 32 - 2 ( x - 1 ) = 12
2 ( x - 1 ) = 32 - 12
2 ( x - 1 ) = 20
x - 1 = 20 : 2
x - 1 = 10
x = 10 + 1
x = 11.
5) 3 ( x + 1 ) - 15 = 36
3 ( x + 1 ) = 36 + 15
3 ( x + 1 ) = 51
x + 1 = 51 : 3
x + 1 = 17
x = 17 - 1
x = 16.
6) 32 - ( 4x - 16 ) = 21
4x - 16 = 32 - 21
4x - 16 = 11
4x = 11 + 16
4x = 27
x = 27 : 4
x = 6,75.
7) 13x - 3x = 500
( 13 - 3 )x = 500
10x = 500
x = 500 : 10
x = 50.
8) 9x + x - 35 = 68
9x + 1x - 35 = 68
9x + 1x = 68 + 35
9x + 1x = 103
( 9 + 1 )x = 103
10x = 103
x = 103 : 10
x = 10,3.
1) 5x = 120 : 2
5x = 60
x = 60 : 5
x = 12.
2) 3x + 27 = 45
3x = 45 - 27
3x = 18
x = 18 : 3
x = 6.
3) 73 + x - 122 = 234 : 2
73 + x - 122 = 117
73 + x = 117 + 122
73 + x = 239
x = 239 - 73
x = 166.
4) 32 - 2 ( x - 1 ) = 12
2 ( x - 1 ) = 32 - 12
2 ( x - 1 ) = 20
x - 1 = 20 : 2
x - 1 = 10
x = 10 + 1
x = 11.
5) 3 ( x + 1 ) - 15 = 36
3 ( x + 1 ) = 36 + 15
3 ( x + 1 ) = 51
x + 1 = 51 : 3
x + 1 = 17
x = 17 - 1
x = 16.
6) 32 - ( 4x - 16 ) = 21
4x - 16 = 32 - 21
4x - 16 = 11
4x = 11 + 16
4x = 27
x = 27 : 4
x = 6,75.
7) 13x - 3x = 500
( 13 - 3 )x = 500
10x = 500
x = 500 : 10
x = 50.
8) 9x + x - 35 = 68
9x + 1x - 35 = 68
9x + 1x = 68 + 35
9x + 1x = 103
( 9 + 1 )x = 103
10x = 103
x = 103 : 10
x = 10,3.
\(\left(x^2-10\right):5=3\\ \Rightarrow x^2-10=3.5\\ \Rightarrow x^2=15+10\\ \Rightarrow x^2=25\\ \Rightarrow\left[{}\begin{matrix}x=-5\\x=5\end{matrix}\right.\)
Lời giải:
a) Xét hiệu:
\(a^4+b^4-(a^3b+ab^3)\)
\(=(a^4-a^3b)-(ab^3-b^4)\)
\(=a^3(a-b)-b^3(a-b)=(a-b)(a^3-b^3)=(a-b)(a-b)(a^2+ab+b^2)\)
\(=(a-b)^2(a^2+ab+b^2)\)
Ta thấy: \((a-b)^2\geq 0, \forall a,b\in\mathbb{R}\)
\(a^2+ab+b^2=(a+\frac{b}{2})^2+\frac{3b^2}{4}\geq 0, \forall a,b\in\mathbb{R}\)
\(\Rightarrow a^4+b^4-(a^3b+ab^3)=(a-b)^2(a^2+ab+b^2)\geq 0, \forall a,b\in\mathbb{R}\)
\(\Rightarrow a^4+b^4\geq ab^3+a^3b\) với mọi $a,b\in\mathbb{R}$
Ta có đpcm.
Dấu "=" xảy ra khi $a=b$
b)
\((x-3)(x-4)(x-5)(x-6)+3\)
\(=[(x-3)(x-6)][(x-4)(x-5)]+3\)
\(=(x^2-9x+18)(x^2-9x+20)+3\)
\(=a(a+2)+3\) (đặt \(x^2-9x+18=a)\)
\(=a^2+2a+3=(a+1)^2+2\geq 2>0, \forall a\in\mathbb{R}\)
hay \((x-3)(x-4)(x-5)(x-6)+3>0, \forall x\in\mathbb{R}\) (đpcm)
a) Xét hiệu:
a4+b4−(a3b+ab3)a4+b4−(a3b+ab3)
=(a4−a3b)−(ab3−b4)=(a4−a3b)−(ab3−b4)
=a3(a−b)−b3(a−b)=(a−b)(a3−b3)=(a−b)(a−b)(a2+ab+b2)=a3(a−b)−b3(a−b)=(a−b)(a3−b3)=(a−b)(a−b)(a2+ab+b2)
=(a−b)2(a2+ab+b2)=(a−b)2(a2+ab+b2)
Ta thấy: (a−b)2≥0,∀a,b∈R(a−b)2≥0,∀a,b∈R
a2+ab+b2=(a+b2)2+3b24≥0,∀a,b∈Ra2+ab+b2=(a+b2)2+3b24≥0,∀a,b∈R
⇒a4+b4−(a3b+ab3)=(a−b)2(a2+ab+b2)≥0,∀a,b∈R⇒a4+b4−(a3b+ab3)=(a−b)2(a2+ab+b2)≥0,∀a,b∈R
⇒a4+b4≥ab3+a3b⇒a4+b4≥ab3+a3b với mọi a,b∈Ra,b∈R
Ta có đpcm.
Dấu "=" xảy ra khi a=ba=b
b)
(x−3)(x−4)(x−5)(x−6)+3(x−3)(x−4)(x−5)(x−6)+3
=[(x−3)(x−6)][(x−4)(x−5)]+3=[(x−3)(x−6)][(x−4)(x−5)]+3
=(x2−9x+18)(x2−9x+20)+3=(x2−9x+18)(x2−9x+20)+3
=a(a+2)+3=a(a+2)+3 (đặt x2−9x+18=a)x2−9x+18=a)
=a2+2a+3=(a+1)2+2≥2>0,∀a∈R=a2+2a+3=(a+1)2+2≥2>0,∀a∈R
hay (x−3)(x−4)(x−5)(x−6)+3>0,∀x∈R(x−3)(x−4)(x−5)(x−6)+3>0,∀x∈R (đpcm)
a) Xét hiệu:
a4+b4−(a3b+ab3)a4+b4−(a3b+ab3)
=(a4−a3b)−(ab3−b4)=(a4−a3b)−(ab3−b4)
=a3(a−b)−b3(a−b)=(a−b)(a3−b3)=(a−b)(a−b)(a2+ab+b2)=a3(a−b)−b3(a−b)=(a−b)(a3−b3)=(a−b)(a−b)(a2+ab+b2)
=(a−b)2(a2+ab+b2)=(a−b)2(a2+ab+b2)
Ta thấy: (a−b)2≥0,∀a,b∈R(a−b)2≥0,∀a,b∈R
a2+ab+b2=(a+b2)2+3b24≥0,∀a,b∈Ra2+ab+b2=(a+b2)2+3b24≥0,∀a,b∈R
⇒a4+b4−(a3b+ab3)=(a−b)2(a2+ab+b2)≥0,∀a,b∈R⇒a4+b4−(a3b+ab3)=(a−b)2(a2+ab+b2)≥0,∀a,b∈R
⇒a4+b4≥ab3+a3b⇒a4+b4≥ab3+a3b với mọi a,b∈Ra,b∈R
Ta có đpcm.
Dấu "=" xảy ra khi a=ba=b
b)
(x−3)(x−4)(x−5)(x−6)+3(x−3)(x−4)(x−5)(x−6)+3
=[(x−3)(x−6)][(x−4)(x−5)]+3=[(x−3)(x−6)][(x−4)(x−5)]+3
=(x2−9x+18)(x2−9x+20)+3=(x2−9x+18)(x2−9x+20)+3
=a(a+2)+3=a(a+2)+3 (đặt x2−9x+18=a)x2−9x+18=a)
=a2+2a+3=(a+1)2+2≥2>0,∀a∈R=a2+2a+3=(a+1)2+2≥2>0,∀a∈R
hay (x−3)(x−4)(x−5)(x−6)+3>0,∀x∈R(x−3)(x−4)(x−5)(x−6)+3>0,∀x∈R (đpcm)v
1, Thực hiện phép tính :
a) \(\left(-37\right)+4\cdot\left|-6\right|\)
\(=-\left(37\right)+4\cdot6\)
\(=-37+24\)
\(=13\)
b) \(17\cdot85+15\cdot17-120\)
\(=17\cdot\left(85+15\right)-120\)
\(=17\cdot100-120\)
\(=1700-120\)
\(=1580\)
x : 4 + x = 3,75
\(x\)x\(\frac{1}{4}\) + \(x=3,75\)
\(x\)x\(\left(\frac{1}{4}+1\right)=\frac{375}{100}\)
\(x\)x\(\frac{5}{4}=\frac{15}{4}\)
\(x=\frac{15}{4}:\frac{5}{4}\)
\(x=\frac{15}{4}\)x\(\frac{4}{5}\)
\(x=3\)
\(\frac{x}{4}+x=3.75\)
\(x\left(\frac{1}{4}+1\right)=\frac{15}{4}\)
\(x\cdot\frac{5}{4}=\frac{15}{4}\)
\(x=\frac{15}{4}:\frac{5}{4}=3\)
a) (3^x-15)^7=0
=) 3x-15=0
=) 3x=15
=) x=5
b) 4^2.x-6=1
4^2.x-6= 4^0 ( vì số nào mũ không cũng bằng 1)
2x-6=0
2x=6
x=3
4.(x-15)=0
=>x-15=0:4=4
x-15=0
=>x=15
4(x-15)=0
(x-15)=0:4
(x-15)=0
x =0+15
x =15